I'll do this with an example, although I don't know how to do the big fancy piecewise bracket curly brace thing:
Consider function $f$ where $$f(x)=\begin{cases} -1 & \text{if }x \leq 0,\\ 1&\text{if }x > 0. \end{cases}$$
And a slight modification of the same function:
$$g(x)=\begin{cases} -1 & \text{if }x < 0,\\ 1&\text{if }x > 0. \end{cases}$$The only difference between $f$ and $g$ is that $0$ is excluded from the domain of $g$. Here is a picture to show both graphs in case that is easier:
My question:
For $f(x)$, since $f(0)$ is defined, we can say $0$ is part of $f$'s domain. However, the two-sided limit does not exist at $x=0$, which means we can say $f$ is discontinuous at $x=0$.
For $g(x)$, we have $x=0$ excluded from the domain. Does this, somewhat counter-intuitively, make $g$ a continuous function? Since the domain is split into something that looks like $(-\infty, 0) \cup (0, \infty)$, each of those two intervals are continuous on their own, and a function is continuous iff it is continuous at all points in its domain.
Is that right? $f$ is discontinuous, but $g$ is continuous?
Yes, $f$ is discontinuous but $g$ is continuous. But the domain of $g$ doesn't just “look like” $(-\infty,0)\cup(0,+\infty)$ (you wrote $\cap$ here, but I suppose that that's a typo); it is that set. By removing from the domain of $f$ the only point at which it was discontinuous, you turned it into a continuous function.